-0.016 738 891 601 562 496 530 15 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 15(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.016 738 891 601 562 496 530 15(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.016 738 891 601 562 496 530 15| = 0.016 738 891 601 562 496 530 15


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.016 738 891 601 562 496 530 15.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.016 738 891 601 562 496 530 15 × 2 = 0 + 0.033 477 783 203 124 993 060 3;
  • 2) 0.033 477 783 203 124 993 060 3 × 2 = 0 + 0.066 955 566 406 249 986 120 6;
  • 3) 0.066 955 566 406 249 986 120 6 × 2 = 0 + 0.133 911 132 812 499 972 241 2;
  • 4) 0.133 911 132 812 499 972 241 2 × 2 = 0 + 0.267 822 265 624 999 944 482 4;
  • 5) 0.267 822 265 624 999 944 482 4 × 2 = 0 + 0.535 644 531 249 999 888 964 8;
  • 6) 0.535 644 531 249 999 888 964 8 × 2 = 1 + 0.071 289 062 499 999 777 929 6;
  • 7) 0.071 289 062 499 999 777 929 6 × 2 = 0 + 0.142 578 124 999 999 555 859 2;
  • 8) 0.142 578 124 999 999 555 859 2 × 2 = 0 + 0.285 156 249 999 999 111 718 4;
  • 9) 0.285 156 249 999 999 111 718 4 × 2 = 0 + 0.570 312 499 999 998 223 436 8;
  • 10) 0.570 312 499 999 998 223 436 8 × 2 = 1 + 0.140 624 999 999 996 446 873 6;
  • 11) 0.140 624 999 999 996 446 873 6 × 2 = 0 + 0.281 249 999 999 992 893 747 2;
  • 12) 0.281 249 999 999 992 893 747 2 × 2 = 0 + 0.562 499 999 999 985 787 494 4;
  • 13) 0.562 499 999 999 985 787 494 4 × 2 = 1 + 0.124 999 999 999 971 574 988 8;
  • 14) 0.124 999 999 999 971 574 988 8 × 2 = 0 + 0.249 999 999 999 943 149 977 6;
  • 15) 0.249 999 999 999 943 149 977 6 × 2 = 0 + 0.499 999 999 999 886 299 955 2;
  • 16) 0.499 999 999 999 886 299 955 2 × 2 = 0 + 0.999 999 999 999 772 599 910 4;
  • 17) 0.999 999 999 999 772 599 910 4 × 2 = 1 + 0.999 999 999 999 545 199 820 8;
  • 18) 0.999 999 999 999 545 199 820 8 × 2 = 1 + 0.999 999 999 999 090 399 641 6;
  • 19) 0.999 999 999 999 090 399 641 6 × 2 = 1 + 0.999 999 999 998 180 799 283 2;
  • 20) 0.999 999 999 998 180 799 283 2 × 2 = 1 + 0.999 999 999 996 361 598 566 4;
  • 21) 0.999 999 999 996 361 598 566 4 × 2 = 1 + 0.999 999 999 992 723 197 132 8;
  • 22) 0.999 999 999 992 723 197 132 8 × 2 = 1 + 0.999 999 999 985 446 394 265 6;
  • 23) 0.999 999 999 985 446 394 265 6 × 2 = 1 + 0.999 999 999 970 892 788 531 2;
  • 24) 0.999 999 999 970 892 788 531 2 × 2 = 1 + 0.999 999 999 941 785 577 062 4;
  • 25) 0.999 999 999 941 785 577 062 4 × 2 = 1 + 0.999 999 999 883 571 154 124 8;
  • 26) 0.999 999 999 883 571 154 124 8 × 2 = 1 + 0.999 999 999 767 142 308 249 6;
  • 27) 0.999 999 999 767 142 308 249 6 × 2 = 1 + 0.999 999 999 534 284 616 499 2;
  • 28) 0.999 999 999 534 284 616 499 2 × 2 = 1 + 0.999 999 999 068 569 232 998 4;
  • 29) 0.999 999 999 068 569 232 998 4 × 2 = 1 + 0.999 999 998 137 138 465 996 8;
  • 30) 0.999 999 998 137 138 465 996 8 × 2 = 1 + 0.999 999 996 274 276 931 993 6;
  • 31) 0.999 999 996 274 276 931 993 6 × 2 = 1 + 0.999 999 992 548 553 863 987 2;
  • 32) 0.999 999 992 548 553 863 987 2 × 2 = 1 + 0.999 999 985 097 107 727 974 4;
  • 33) 0.999 999 985 097 107 727 974 4 × 2 = 1 + 0.999 999 970 194 215 455 948 8;
  • 34) 0.999 999 970 194 215 455 948 8 × 2 = 1 + 0.999 999 940 388 430 911 897 6;
  • 35) 0.999 999 940 388 430 911 897 6 × 2 = 1 + 0.999 999 880 776 861 823 795 2;
  • 36) 0.999 999 880 776 861 823 795 2 × 2 = 1 + 0.999 999 761 553 723 647 590 4;
  • 37) 0.999 999 761 553 723 647 590 4 × 2 = 1 + 0.999 999 523 107 447 295 180 8;
  • 38) 0.999 999 523 107 447 295 180 8 × 2 = 1 + 0.999 999 046 214 894 590 361 6;
  • 39) 0.999 999 046 214 894 590 361 6 × 2 = 1 + 0.999 998 092 429 789 180 723 2;
  • 40) 0.999 998 092 429 789 180 723 2 × 2 = 1 + 0.999 996 184 859 578 361 446 4;
  • 41) 0.999 996 184 859 578 361 446 4 × 2 = 1 + 0.999 992 369 719 156 722 892 8;
  • 42) 0.999 992 369 719 156 722 892 8 × 2 = 1 + 0.999 984 739 438 313 445 785 6;
  • 43) 0.999 984 739 438 313 445 785 6 × 2 = 1 + 0.999 969 478 876 626 891 571 2;
  • 44) 0.999 969 478 876 626 891 571 2 × 2 = 1 + 0.999 938 957 753 253 783 142 4;
  • 45) 0.999 938 957 753 253 783 142 4 × 2 = 1 + 0.999 877 915 506 507 566 284 8;
  • 46) 0.999 877 915 506 507 566 284 8 × 2 = 1 + 0.999 755 831 013 015 132 569 6;
  • 47) 0.999 755 831 013 015 132 569 6 × 2 = 1 + 0.999 511 662 026 030 265 139 2;
  • 48) 0.999 511 662 026 030 265 139 2 × 2 = 1 + 0.999 023 324 052 060 530 278 4;
  • 49) 0.999 023 324 052 060 530 278 4 × 2 = 1 + 0.998 046 648 104 121 060 556 8;
  • 50) 0.998 046 648 104 121 060 556 8 × 2 = 1 + 0.996 093 296 208 242 121 113 6;
  • 51) 0.996 093 296 208 242 121 113 6 × 2 = 1 + 0.992 186 592 416 484 242 227 2;
  • 52) 0.992 186 592 416 484 242 227 2 × 2 = 1 + 0.984 373 184 832 968 484 454 4;
  • 53) 0.984 373 184 832 968 484 454 4 × 2 = 1 + 0.968 746 369 665 936 968 908 8;
  • 54) 0.968 746 369 665 936 968 908 8 × 2 = 1 + 0.937 492 739 331 873 937 817 6;
  • 55) 0.937 492 739 331 873 937 817 6 × 2 = 1 + 0.874 985 478 663 747 875 635 2;
  • 56) 0.874 985 478 663 747 875 635 2 × 2 = 1 + 0.749 970 957 327 495 751 270 4;
  • 57) 0.749 970 957 327 495 751 270 4 × 2 = 1 + 0.499 941 914 654 991 502 540 8;
  • 58) 0.499 941 914 654 991 502 540 8 × 2 = 0 + 0.999 883 829 309 983 005 081 6;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.016 738 891 601 562 496 530 15(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 15(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 530 15(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(2) × 2-6


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1017(10) =

011 1111 1001(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


Decimal number -0.016 738 891 601 562 496 530 15 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100